CHAPTER 5 Structural Analysis

STATICS

Chapter 5 – Structural Chapter 5 – Structural AnalysisAnalysis

Learning Objectives

• Define a truss and identify its members.

• Compute the internal forces in structural elements using the method of joints and the method of sections.

• Identify zero-force members.

• Analyze frames and machines.

WHAT IS ANALYSIS OF STRUCTURES?

A STRUCTURE’S PURPOSE IS TO TRANSMIT APPLIED LOADS THROUGH THE STRUCTURE TO IT’S EXTERNAL SUPPORTS

THE STRUCTURE IS A FRAME WHERE INTERNAL MEMBERS

ARE SUBJECTED TO BOTH AXIAL AND TRANSVERSE LOADS

STRUCTURAL ANALYSIS IS USED TO DETERMINE THE FORCE OR FORCES THAT EACH PART OF THE STRUCTURE MUST SUPPORT

The Eiffel Tower

• The Eiffel Tower is a truss structure comprising more than 15,000 girders (pieces of iron) and 2,500,000 rivets.

• For its height of 1,063 feet, it uses a meager 7,300 tons of iron.

The Eiffel Tower• To appreciate the power of

trusses, let us compare the Eiffel Tower with the Washington Monument whose height is 555 feet and weight is 82,000 tons.

• Even though it is made of different materials and reaches to twice the height, the Eiffel Tower uses substantially less material – a full order of magnitude less material, in fact – than the Washington Monument does.

• The contrast shows the advantage in structural efficiency that a truss structure enjoys over a solid beam.

555 ft82,000 tons

1,063 ft7,300 tons

• For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered.

• In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

• Three categories of engineering structures are considered:

a) Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces.

b) Trusses: formed from two-force members, i.e., straight members with end point connections

c) Machines: structures containing moving parts designed to transmit and modify forces.

Truss• Often, we encounter the problem of supporting external

loads from a distance.

• The bridge transfers the weight of cars/trucks, self-weight, and wind loads to the support structures and to the banks of the river.

• A truss is a structural frame consisting of straight members, all lying in the same plane, connected to form a triangle or series of triangles.

Definition of a Truss

• A truss consists of straight members connected at joints. No member is continuous through a joint.

• Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two-force members are considered.

• Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure.

• When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.


Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints.

Truss• We can achieve the purpose of transferring load to the support

structure in a number of different ways.

• One possible design is to join several slender structural elements using pinned connections. We refer to such a structure as a truss.

• The slender members known as struts intersect at truss joints.

• The struts forming the top and bottom of the truss are referred to as chords.

• The inclined and vertical elements connecting the upper and lower chords are known as the web of the truss.

• Typically, chords carry significantly higher loads than the web elements.

Struts

• The struts are two-force members i.e. only two forces act on the strut. To satisfy the force equilibrium equation with only two forces acting, these forces must be equal and opposite in direction.

• To satisfy the moment equilibrium equation, the forces must be collinear.

• Depending on their direction, the forces subject the struts to compression or tension (pushing or pulling).

Truss

• A truss is a structure made of slender structural elements that take loads only in the form of compression or tension.

• At a fundamental level, a truss must support the load and remain stable.

• When the load is applied, the truss should not collapse, and its struts should not move.

Development of a Truss• Most trusses build on the

concept of a triangle, as the triangle is inherently the most stable geometric form.

• We cannot change the shape of a triangle without distorting one of its sides.

• The pin support provides force reactions in the horizontal and vertical directions.

• The roller support provides a vertical reaction.

• The truss cannot move in any direction.

Development of Truss

• Tetrahedrons and octahedrons are the most stable three-dimensional shapes, offering more stability than a cube.

• On closer inspection, one will notice that a tetrahedron and an octahedron are, in fact, an assemblage of triangles.

Space Trusses

• An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron.

• A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time.

• Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations.

• In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints.

• Conditions of equilibrium for the joints provide 3n equations. For a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions.

Truss

• Other geometric forms change their shapes when the load is applied.

• A rectangle transforms into a parallelogram without its sides distorting.

Simple Truss

• To prevent the collapse of this truss, we can add a diagonal strut.

• We can view the new truss as one that is made of two triangles.

• We can synthesize planar trusses (ones that lie in a single plane) by putting triangular trusses together.

• A planar truss made of triangles is called a simple truss.

Assumptions in Truss Analysis

• All loads are applied at the joints: – The loads are applied only at the joints. – Neglect the self-weight of the struts. If the self-

weight is significant in comparison to the external loads, then apply half of the weight of each strut at its corresponding joint.

• The joints do not transmit moment loads: – The struts in a truss comprise two force members

which accept loads as compression or tension. – The joints do not to transmit moment loads. A

smooth pin connection satisfies this assumption.

Method of Joints

• The name for this method derives from its procedure: analyze one joint at a time by examining the forces that the struts exert on the joint.

• This procedure involves the following steps:– Solve unknown external reactions– Analyze one-joint at a time

Method of Joints

• Solve unknown external reactions:

– Draw the free-body diagram for the entire structure.

– Write the equilibrium equations.

– Solve for the support reactions.

Method of Joints

• Analyze one-joint at a time: – Choose a joint: Select a joint with a maximum of two unknown

forces and a minimum of one known force.

– Draw the free-body diagram for the joint: While drawing the free-body diagram, point the unknown forces away from the joint. In other words, the strut is under tension and therefore pulls the joint.

– Write the force equilibrium equations: As all the forces pass through the joint, we only establish the force equilibrium equation. It is not useful to establish the moment equilibrium equation as it does not provide any insight.

– Solve for the forces exerted by the struts. If we get a positive value for the unknown force, then the strut is experiencing tension and the direction shown in the free body diagram is correct. A negative value indicates that the strut is experiencing compression. The direction of the vector should be switched.

Example 1

• The Warren truss is the most common truss for bridge construction.

• For this truss structure, determine the forces in the members of the truss.

Example 1• Step 1: Solve unknown

external reactions– Draw the free-body diagram

for the entire structure

– Write the equilibrium equations

– Solve for the support reactions

0

1000 1000 1000 0

1000lb 3ft 1000lb 6ft 1000 lb 9 ft 12ft =0

x Ax

y Ay By

By

F R

F R R

RAM

1500 lbByR

1500 lbAyR

The results make sense because the loading is symmetric and, therefore, equally distributed on joints A and B.

Example 1

cos 45 0

sin 45 1500 0

x

y

F

F

AC AD

AC

F F

F

cos 45 0

sin 45 1500 0

x

y

F

F

BE BD

BE

F F

F

15002120 lb

sin 451500 lb

AC

AD

F

F

15002120 lb

sin 451500 lb

BE

BD

F

F

Choose a joint Joint A Joint B

Draw the free-body diagram

Write the force equilibrium equations

Solve for the forces exerted by

the struts

Example 1

• Points to consider:– The forces in the web members AC and BE are

negative. In other words, the force in these members manifests as compression, not tension.

– The forces in struts AC and BE are equal, as are the forces in struts AD and BD. The same conclusion can be reached without analyzing joint B by looking at the effect of symmetry.

– Now, we analyze joint C, solve for the force in strut CD, and then use the effect of symmetry to identify the forces in the rest of the structure.

Example 1

2120cos 45 cos 45 0

2120sin 45 1000 sin 45 0

x

y

F

F

CE CD

CD

F F

F

707 lb

2000 lb

CD

CE

F

F

Choose a joint Joint C

Draw the free-body diagram

Establish the force equilibrium equations

Solve for the forces exerted by the struts

Example 1

Determine the loads in each member if the magnitude of force P is 1000 lb.

0

4 ft 8 ft 0

500 lb20

500 lb2

x

A

y

F

M

F

Ax

Ax

By

By

By Ay

Ay

P R

R P

P R

PR

R R

PR

Joint A0

0

0

sin 45 0

707 lbsin 45

cos45 0

500 lb

y

x

F

F

AC Ay

AyAC

AB AC Ax

AB

F R

RF

F F R

F =

Joint C0 0cos45 cos45 0

707 lb

yF CA CB

CB CA

F F

F F

Zero-Force Members• Definition: A zero-force member

is a strut that does not carry any load.

• These zero-force members serve three purposes:

– Prevent buckling in the long structural members by increasing the rigidity in the transverse direction.

– Become redundant members and come into play when normally load-carrying structural elements fail.

– Support loads during construction.

Zero-Force Members

• Let us draw the free-body diagram for the node G.

• Two forces FGC and FGE are collinear and oppose each other.

• Force FGD remains unbalanced. To achieve equilibrium, FGD must be equal to zero.

Zero-Force Members

• Let us analyze joint F by drawing.

• Two non-collinear forces are acting at this joint.

• These forces will satisfy the equilibrium equations only if the magnitude of both forces is equal to zero.

Zero-Force Members• We can generalize these two observations and formulate our two rules to

identify zero-force members:

– At a joint where three members meet, two of the members are collinear, and if there is no external load, the non-collinear member is a zero-force member.

• At a joint where two members meet and there is no external load, the two members are zero-force members.

• Note that to identify zero-force members, we should look at the joints that bear no external load.


Note the member CD is a zero-force member.The solution is same as before.


0

500 lb from symmetry2

xF Ax

Ay By

R

PR R

Joint A

0

0

0

sin 45 500 lb 0

500 lb 707 lbsin 45

cos45 0

500 lb

y

x

F

F

AC

AC

AC AD

AD

F

F

F F

F =

Joint AJoint D

0

1000 lbCD

yF CDP F

F P

Method of Sections

• As the name implies, the method of sections uses an imaginary sectional plane to divide the truss into two parts.

• Because the entire truss is in equilibrium, each of its parts should also be in equilibrium.

• We can draw the free-body diagram for one of the two parts of the truss and set the equilibrium equations.

Method of Sections

• As we can establish three equilibrium equations for a planar structure, we can solve for a maximum of three unknown forces.

• The sectioning plane must not cut across more than three struts with unknown forces.

• This method is very useful in finding the forces in a certain member in a large truss without determining the forces in all the individual struts.

Method of Sections

• The procedure for the method of sections involves two basic steps:

– Solve unknown external reactions

– Analyze a part of the truss

Method of Sections

• Solve unknown external reactions: Note that by, depending on the type of structure, we may be able to solve the problem even by skipping this step.

– Draw the free-body diagram for the entire structure.


– Solve for the support reactions.

Method of Sections

• Analyze a part of the truss:

– Choose a sectional plane: Create a sectional plane that cuts across a maximum of three struts with unknown forces.

– Draw the free-body diagram for a part of the truss: Pick a part of the truss which is easy to analyze. While drawing the free-body diagram, point out the unknown forces pulling the strut. In other words, the strut is experiencing tension.


– Solve for the forces exerted by the struts. If we get a positive value for the unknown force, then the strut is in tension and the direction shown in the free body diagram is correct. A negative value indicates that the strut is in compression. The direction of the vector should be reversed.

Example 2

• Determine the forces in struts EG, FG, and FH.

• This truss is known as the Pratt truss.

• In this truss, all the diagonal web elements except the end elements are slanted towards the center.

Example 2

• STEP 1: Determine the support reactions.– By looking at the

symmetry, we can write the support reactions as

0

5 kN

5 kN

Ax

Ay

By

R

R

R

Example 2

• STEP 2: Analyze a part of the truss– Choose a sectional

plane: The first step involves the creation of a sectional plane.

– Interested in finding the forces in struts EG, FG, and FH.

– Draw a vertical section cutting these three struts.

Example 2

• Draw the free-body diagram for a part of the truss: – The left side is easier

to solve because it involves fewer loads.

– As we do not know the sense of the forces in struts EG, FG, and FH, we can assume that they are pulling the struts.

Example 2

• Write the equilibrium equations:

– An appropriate choice of the point about which moments are taken can greatly simplify the problem.

– If we take the moments about point F, the moment equation will involve the force in strut EG only. The forces in struts FH and FG will not produce any moment about the point F.

sin 45 0

5 kN 2 kN 1 kN cos 45 0

5 kN 4 m 2 kN 2 m 2 m 0

x

y

F

F

EG FG FH

FG

F EG

F F F

F

M F

Example 2

• Solve for the forces exerted by the struts.

8kN

2.83kN

10kN

EG

FG

FH

F

F

F

Frames and Machines

• Frames are stationary structures that comprise pinned structural members carrying multiple forces and moments.

• Machines, on the other hand, contain moving parts along with stationary members.

• The analysis of frames and machines is more complex because the method for solving the problem is more flexible, requiring more thinking on our part to structure the solution.

Frames and Machines

• The general procedure for analyzing frames or machines involves two steps.

– Step I - Apply the equilibrium equations to the entire structure

– Step II – Apply the equilibrium equations to individual members or groups of members

Isolate each part by drawing its outlined shape. Then show all the forces and/or couple moments that act on the part.Identify all the two-force members in the structure and represent their free-body diagrams as having two equal but opposite collinear forces acting at their points of application. - Forces common to any two contacting members act with equal magnitudes but opposite sense on the respective members.

Frames and Machines

• Step I - Apply the equilibrium equations to the entire structure:

– While we determined the reaction forces using the equilibrium equations in the case of trusses, often we may not be able to find the reaction forces in frames and machines using the equilibrium equations of the entire structure alone.

– Make an effort to determine as many reactions as possible.

Frames and Machines• Step II – Apply the equilibrium equations to individual members or

groups of members:

– Disassemble the structure into parts composed of individual members or groups of members.

– Draw the free-body diagrams for the parts. Represent only the external forces.

– When drawing a free-body diagram for individual components, identify the forces acting between components and make sure to label them to be equal and opposite. Assume the direction of unknown forces.

– Try to identify the two force members and represent the forces along the length of the member.

– Apply the equilibrium equations and try to solve from a member with the least number of unknowns. Sometimes, in complex problems, it becomes necessary to solve the equations simultaneously.

Example 3

• Determine the loads carried by each member of the two-member frame.

Example 3

• Step I - Apply the equilibrium equations to the entire structure– Draw the free-body

diagram for the entire structure.

– Support A provides reactions in the x- and y-directions, so we identify two forces, Ax and Ay.

– The fixed support C provides both force and moment reactions.

Example 3

• Step I - Apply the equilibrium equations to the entire structure

0

200lb 0

10ft 200lb 3ft 0

x Ax Cx

y Ay Cy

Ay

F R R

F R R

RCM

Ax CxR R 60lbAyR

140lbCyR

Ax CxR R 60lbAyR

ΣMB = RAy (4) + RAx(4)=0

RAx = 4RAy

4=-60 lb

RA

-60

-60

85.84)6060( 22 A

R

52.6445cos85.84

45cos

CX

CXBX

R

RFF

Example 4

• Determine the loads in the structural members.

Frame – Example Frame – Example ProblemProblem

The hoist frame is used to lift the weight W. The pull T on the pulley balances the weight W. For W= 4 kN, determine all forces acting on sketches of the members. Assume that the weight of the members are small compared to the load and the wall is smooth.


Look at the pulley system, the tension in the wire is 4 kN. Therefore

4 kN 4 kN 8 kN

P T W

P


Using equilibrium to solve for the points

x Ax E

y Ay

Ay

A E

E

Ax

8 kN 0

8 kN

8 kN 4 m 3 m

10.67 kN

10.67 kN

F R R

F R

R

M R

R

R


Look at member ABC

x x

x

y y AB

C AB

AB

y

10.67 kN 0

10.67 kN

8 kN 8 kN 0

8 kN 4 m 2 m

16 kN

16.0 kN

F C

C

F C T

M T

T

C

Mechanism

A mechanism is a non-rigid mechanical system whose components are capable of relative motion when not attached to a support.

Mechanism – Example Mechanism – Example ProblemProblem

A tackle on the football team weighs 350 lb. In coming off his stance, he places his entire weight on the ball of his foot. In effect, his foot acts as a lever. The weight of the tackle is transmitted to his foot through the tibia. In tern the Achilles tendon pulls on the heal consider the static effects only, determine the force that the tendon must withstand.

Mechanism – Example Mechanism – Example ProblemProblem

Do a free-body diagram of the foot

x x

x

y y

F

y

0

0 lb

350 lb 0

350 lb 12 in 14 in

300 lb

50 lb

F F

F

F F T

M T

T

F

Example – Machine Example – Machine ProblemProblem

In using the bolt cutter, a worker applies two 100-lb forces to the handles. Determine the magnitude of the forces exerted by the cutters at the bolt.


Explode the head off of the shears:

and continue

RB can be solve using the moment at C


Find the resultant force at B

C B

B

100 lb 18.5 in 0.5 in

3700 lb

M R

R


Look at the head of the shears

A w

w

3700 lb 4 in 1.0 in

14800 lb

M R

R

Example 5

• Determine the required weight W to keep the system in equilibrium.

Example 5

• Draw the free-body diagrams for the three pulleys and two weights.

• Except for the 10 lb weight, all other forces are unknowns.

• The dotted lines show the action-reaction pairs acting between different objects.

Example 5• Now, we can solve by using the

equilibrium equation for the 10 lb weight. We get

• Applying the equilibrium equations for pulley C, we get

• We can apply the equilibrium equations to pulley A. We get

• Applying the equilibrium equation to the unknown weight, we get

10 lb 0

10 lb

2

2

F T

T

2 0

2 20 lb

1 2

1 2

F T T

T T

2 0

2 40 lb

1 1

1 1

F T F

F T

0

50 lb

1 2

1 2

F F T W

W F T

CHAPTER 5 Structural Analysis

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